What Is an Inequality?
An inequality is a mathematical statement that compares two expressions using inequality symbols: less than (<), greater than (>), less than or equal to (<=), or greater than or equal to (>=). Unlike an equation, which has a specific solution (like x = 5), an inequality has a range of solutions (like x > 5, meaning any number greater than 5).
Inequalities describe real-world constraints all the time. If you need at least 70 points to pass a test, that is an inequality: score >= 70. If a bridge can hold at most 10 tons, that is an inequality: weight <= 10. If you need more than $50 to buy a jacket, that is: money > 50.
Solving an inequality means finding all values of the variable that make the inequality true. The solution set is typically expressed as an inequality (x > 3), in interval notation ((3, infinity)), or graphically on a number line. The process of solving is very similar to solving equations, with one critical difference: multiplying or dividing both sides by a negative number reverses the inequality sign.
Solving One-Step and Two-Step Linear Inequalities
One-step inequalities are solved exactly like one-step equations. For x + 4 > 9: subtract 4 from both sides to get x > 5. For 3x <= 15: divide both sides by 3 to get x <= 5. For x/2 >= 7: multiply both sides by 2 to get x >= 14.
Two-step inequalities require two operations. For 2x + 3 < 11: subtract 3 from both sides (2x < 8), then divide by 2 (x < 4). For (x - 5)/3 >= 2: multiply both sides by 3 (x - 5 >= 6), then add 5 (x >= 11).
The critical rule: when you multiply or divide both sides of an inequality by a negative number, you must flip the inequality sign. For -3x > 12: divide both sides by -3 AND flip the sign: x < -4. Why? Because multiplying by a negative number reverses the order of numbers on the number line. Since -3(5) = -15 and -3(4) = -12, and -15 < -12, the "greater" relationship flips.
This is the most common mistake students make with inequalities. Always check: did I multiply or divide by a negative number? If yes, flip the sign. If no, leave it as is. You can verify your answer by plugging in a test value from your solution set. For x < -4, try x = -5: -3(-5) = 15 > 12. Correct.
Multi-Step Linear Inequalities
Multi-step inequalities follow the same strategy as multi-step equations: simplify each side, then isolate the variable. Example: Solve 4(x - 2) + 3 > 2x + 7.
Step 1 — Distribute: 4x - 8 + 3 > 2x + 7. Step 2 — Combine like terms: 4x - 5 > 2x + 7. Step 3 — Move variable terms to one side: subtract 2x from both sides: 2x - 5 > 7. Step 4 — Isolate x: add 5 to both sides: 2x > 12. Divide by 2: x > 6.
When variables appear on both sides, keep the variable positive to avoid flipping the sign unnecessarily. For 3 - 2x < 5x + 10: instead of subtracting 5x from both sides (which gives -7x < 10, requiring a flip), add 2x to both sides (3 < 7x + 10), then subtract 10 (-7 < 7x), then divide by 7 (-1 < x, or x > -1). Same answer, fewer opportunities for sign errors.
Check your work by testing values on both sides of the boundary. For x > 6, test x = 7 (should be true) and x = 5 (should be false). Plug into the original inequality: 4(7 - 2) + 3 = 23. 2(7) + 7 = 21. Is 23 > 21? Yes. Now test x = 5: 4(3) + 3 = 15. 2(5) + 7 = 17. Is 15 > 17? No. This confirms x > 6 is correct.
Compound Inequalities
A compound inequality combines two inequalities with "and" or "or." An "and" compound inequality (also written as a double inequality) requires both conditions to be true simultaneously. An "or" compound inequality requires at least one condition to be true.
"And" example: Solve 3 < 2x + 1 < 9. This means 2x + 1 > 3 AND 2x + 1 < 9. Solve the double inequality by applying the same operation to all three parts. Subtract 1: 2 < 2x < 8. Divide by 2: 1 < x < 4. The solution is all values of x between 1 and 4 (not including 1 and 4). In interval notation: (1, 4).
"Or" example: Solve 3x - 1 < -7 OR 2x + 3 > 11. Solve each inequality separately. First: 3x < -6, so x < -2. Second: 2x > 8, so x > 4. The solution is x < -2 or x > 4. In interval notation: (-infinity, -2) union (4, infinity). On a number line, you shade everything left of -2 and everything right of 4, with the region between -2 and 4 unshaded.
Graphing compound inequalities: "and" inequalities produce a segment (values between two numbers). "Or" inequalities produce two rays going in opposite directions. Use open circles for < and >, closed circles for <= and >=.
Absolute Value Inequalities
Absolute value measures the distance of a number from zero. The inequality |x| < 3 means "the distance from x to zero is less than 3," which gives -3 < x < 3. The inequality |x| > 3 means "the distance from x to zero is greater than 3," which gives x < -3 or x > 3.
Rule for |expression| < k (or <=): rewrite as -k < expression < k (a compound "and" inequality). Example: |2x - 5| < 7. Rewrite: -7 < 2x - 5 < 7. Add 5: -2 < 2x < 12. Divide by 2: -1 < x < 6.
Rule for |expression| > k (or >=): rewrite as expression < -k OR expression > k (a compound "or" inequality). Example: |3x + 1| >= 10. Rewrite: 3x + 1 <= -10 OR 3x + 1 >= 10. Solve each: 3x <= -11, so x <= -11/3. And 3x >= 9, so x >= 3. Solution: x <= -11/3 or x >= 3.
Special cases: |expression| < 0 has no solution (absolute value is always >= 0). |expression| > 0 is true for all values except where the expression equals 0. |expression| <= 0 has exactly one solution: where the expression equals 0. |expression| >= 0 is true for all real numbers.
Quadratic Inequalities
A quadratic inequality looks like ax^2 + bx + c > 0 (or <, <=, >=). To solve one, first find the zeros of the corresponding equation ax^2 + bx + c = 0. These zeros divide the number line into intervals. Then test a value from each interval to determine where the inequality is satisfied.
Example: Solve x^2 - 5x + 6 > 0. Factor: (x - 2)(x - 3) > 0. The zeros are x = 2 and x = 3. They divide the number line into three intervals: (-infinity, 2), (2, 3), and (3, infinity). Test a value from each: x = 0: (0 - 2)(0 - 3) = 6 > 0 (true). x = 2.5: (0.5)(-0.5) = -0.25 > 0 (false). x = 4: (2)(1) = 2 > 0 (true). Solution: x < 2 or x > 3.
For x^2 - 5x + 6 <= 0, the solution is the opposite interval (where the quadratic is negative or zero): 2 <= x <= 3. Include the endpoints because the inequality allows equality.
Sign analysis shortcut: for (x - a)(x - b) > 0 where a < b, the product is positive when both factors are positive (x > b) or both are negative (x < a). The product is negative when the factors have opposite signs (a < x < b). This pattern — positive/negative/positive — always holds for a standard upward-opening parabola with two real roots.
Graphing Inequalities on a Number Line
Graphing inequalities on a number line is a visual way to represent the solution set. The conventions are: use an open circle (or parenthesis) for strict inequalities (< and >) and a closed circle (or bracket) for inclusive inequalities (<= and >=). Shade the region of the number line that contains the solutions.
For x > 3: draw an open circle at 3 and shade everything to the right. For x <= -2: draw a closed circle at -2 and shade everything to the left. For 1 < x <= 5: draw an open circle at 1, a closed circle at 5, and shade the region between them.
For compound inequalities: "and" compounds produce a shaded segment between two points. "Or" compounds produce two separate shaded rays. For quadratic inequalities, the solution may be one segment or two rays, depending on the direction of the inequality.
Interval notation provides a compact written alternative to graphing. Use parentheses for open endpoints and brackets for closed endpoints. (3, infinity) means x > 3. [-2, 5) means -2 <= x < 5. (-infinity, -1) union (4, infinity) means x < -1 or x > 4. Infinity always gets a parenthesis because it is not a specific number and cannot be included.
Practice Problems
Problem 1: Solve 5x - 3 >= 2x + 9. Solution: Subtract 2x: 3x - 3 >= 9. Add 3: 3x >= 12. Divide by 3: x >= 4.
Problem 2: Solve -4x + 1 > 13. Solution: Subtract 1: -4x > 12. Divide by -4 AND flip: x < -3. Verify: test x = -4: -4(-4) + 1 = 17 > 13. True.
Problem 3: Solve |x - 4| <= 6. Solution: Rewrite as -6 <= x - 4 <= 6. Add 4: -2 <= x <= 10. Interval notation: [-2, 10].
Problem 4: Solve x^2 - x - 12 < 0. Factor: (x - 4)(x + 3) < 0. Zeros: x = 4 and x = -3. Test intervals: x = -4: (-8)(1) = -8 < 0 (false — wait, (-4-4)(-4+3) = (-8)(-1) = 8 > 0, false). x = 0: (0-4)(0+3) = -12 < 0 (true). x = 5: (1)(8) = 8 > 0 (false). Solution: -3 < x < 4. If any of these inequality types are giving you trouble, take a photo of the problem and send it to ScanSolve for a step-by-step walkthrough.
